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dimulka [17.4K]
3 years ago
9

If your computers normal zoom is 100, but it looks like its at 500, how do you stop that? my invisioned thing isn't on, neither

is an accessebility. and right now, everything looks huge, I have to put the zoom on 50.
Computers and Technology
1 answer:
Alex Ar [27]3 years ago
7 0

Answer:

You just need to click over the desktop, move to the View, and then change to medium, if its the icon size.

However, if you feel your resolution requires screwing up, then you are required to do the right-click, and again over the Desktop and then move to the settings (screen resolution),

And if you by chance clicked on the magnifier utility, you can CTRL + ALT +DEL and you will also be required to shut the interface.

Explanation:

You just need to click over the desktop, move to the View, and then change to medium, if its the icon size.

However, if you feel your resolution requires screwing up, then you are required to do the right-click, and again over the Desktop and then move to the settings (screen resolution),

And if you by chance clicked on the magnifier utility, you can CTRL + ALT +DEL and you will also be required to shut the interface.

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salantis [7]

Google Analytics, this is a common one but there are others that do the same thing.

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3 years ago
class student_record { public: int age; string name; double gpa; }; Implement a void function that has five formal parameters: a
Setler79 [48]

Answer:

void delete_record(student_record *arr, int &size, int age, string name, double gpa) {

int index = -1;

if (arr != NULL && size > 0) {

for (int i = 0; i < size; ++i) {

if (arr[i].age == age && arr[i].name == name && arr[i].gpa == gpa) {

index = i;

break;

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}

}

if (index != -1) {

for (int i = index; i < size - 1; ++i) {

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8 0
3 years ago
R15. Suppose there are three routers between a source host and a destination host. Ignoring fragmentation, an IP datagram sent f
maria [59]

Answer:

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8 0
3 years ago
The area or the window that shows the current folder location​
Alex Ar [27]

Answer:

The focused window

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8 0
3 years ago
The compare_strings function is supposed to compare just the alphanumeric content of two strings, ignoring upper vs lower case a
Korolek [52]

Answer:

There is a problem in the given code in the following statement:

Problem:

punctuation = r"[.?!,;:-']"

This produces the following error:

Error:

bad character range

Fix:

The hyphen - should be placed at the start or end of punctuation characters. Here the role of hyphen is to determine the range of characters. Another way is to escape the hyphen - using using backslash \ symbol.

So the above statement becomes:

punctuation = r"[-.?!,;:']"  

You can also do this:

punctuation = r"[.?!,;:'-]"  

You can also change this statement as:

punctuation = r"[.?!,;:\-']"

Explanation:

The complete program is as follows. I have added a print statement print('string1:',string1,'\nstring2:',string2) that prints the string1 and string2 followed by return string1 == string2  which either returns true or false. However you can omit this print('string1:',string1,'\nstring2:',string2) statement and the output will just display either true or false

import re  #to use regular expressions

def compare_strings(string1, string2):  #function compare_strings that takes two strings as argument and compares them

   string1 = string1.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   string2 = string2.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   punctuation = r"[-.?!,;:']"  #regular expression for punctuation characters

   string1 = re.sub(punctuation, r"", string1)  # specifies RE pattern i.e. punctuation in the 1st argument, new string r in 2nd argument, and a string to be handle i.e. string1 in the 3rd argument

   string2 = re.sub(punctuation, r"", string2)  # same as above statement but works on string2 as 3rd argument

   print('string1:',string1,'\nstring2:',string2)  #prints both the strings separated with a new line

   return string1 == string2  # compares strings and returns true if they matched else false

#function calls to test the working of the above function compare_strings

print(compare_strings("Have a Great Day!","Have a great day?")) # True

print(compare_strings("It's raining again.","its raining, again")) # True

print(compare_strings("Learn to count: 1, 2, 3.","Learn to count: one, two, three.")) # False

print(compare_strings("They found some body.","They found somebody.")) # False

The screenshot of the program along with its output is attached.

4 0
3 years ago
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